la forma correcta de utilizar la ecuaciÓn de bernoulli · however, if the fluid flow is...

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LA FORMA CORRECTA DE UTILIZAR LA ECUACIÓN DE BERNOULLI RESUMEN Cuando se pretende aplicar la ecuación de Bernoulli en un caso determinado de un problema de flujo de fluido, se deben cumplir algunas restricciones para aplicar correctamente esta ecuación en particular. El flujo de fluido debe considerarse no viscoso, incompresible, estable e irrotacional. Sin embargo, si el flujo de fluido es rotacional, la ecuación de Bernoulli aún puede aplicarse siempre y cuando los puntos de interés estén en la misma línea de corriente del flujo. Aquí, nos enfocaremos en demostrar que, en el caso de un flujo de fluido rotacional, los puntos de interés deben encontrarse en la misma línea de corriente del flujo y por esta razón, sí se podría continuar con el uso de la ecuación de Bernoulli. El principio solo es aplicable a los flujos isentrópicos: cuando los efectos de los procesos irreversibles (e.g. turbulencia, fricción) y los procesos no adiabáticos (e.g. la radiación de calor, difusión de masa) son pequeños y pueden despreciarse. Palabras clave: flujo, Bernoulli, viscoso, incompresible, estable, irrotacional, línea de corriente, isentrópico, irreversible, turbulencia, fricción

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Page 1: LA FORMA CORRECTA DE UTILIZAR LA ECUACIÓN DE BERNOULLI · However, if the fluid flow is rotational, Bernoulli`s equation can still be applicable as long as the points of interest

LA FORMA CORRECTA DE UTILIZAR LA ECUACIÓN DE BERNOULLI

RESUMEN

Cuando se pretende aplicar la ecuación de Bernoulli en un caso determinado de un

problema de flujo de fluido, se deben cumplir algunas restricciones para aplicar

correctamente esta ecuación en particular. El flujo de fluido debe considerarse no

viscoso, incompresible, estable e irrotacional. Sin embargo, si el flujo de fluido es

rotacional, la ecuación de Bernoulli aún puede aplicarse siempre y cuando los

puntos de interés estén en la misma línea de corriente del flujo. Aquí, nos

enfocaremos en demostrar que, en el caso de un flujo de fluido rotacional, los puntos

de interés deben encontrarse en la misma línea de corriente del flujo y por esta

razón, sí se podría continuar con el uso de la ecuación de Bernoulli. El principio solo

es aplicable a los flujos isentrópicos: cuando los efectos de los procesos

irreversibles (e.g. turbulencia, fricción) y los procesos no adiabáticos (e.g. la

radiación de calor, difusión de masa) son pequeños y pueden despreciarse.

Palabras clave: flujo, Bernoulli, viscoso, incompresible, estable, irrotacional, línea

de corriente, isentrópico, irreversible, turbulencia, fricción

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THE CORRECT WAY TO USE BERNOULLI`S EQUATION

ABSTRACT

When Bernoulli's equation is intended to apply on a certain case of a fluid flow

problem, some restrictions must be met in order to correctly apply this particular

equation. The fluid flow must be considered inviscid, incompressible, steady and

irrotational. However, if the fluid flow is rotational, Bernoulli`s equation can still be

applicable as long as the points of interest are on the same streamline of the fluid

flow. Here, we will focus on demonstrate that, in the case of a rotational fluid flow,

the points of interest must be on the same streamline and because of that, it can be

proceeding with the usage of Bernoulli`s equation. The principle is only applicable

for isentropic fluid flows: when the effects of irreversible processes (e.g. turbulence,

friction) and non-adiabatic processes (e.g. heat radiation, mass diffusion) are small

and can be neglected.

Key words: fluid, Bernoulli, inviscid flow, incompressible, steady, irrotational,

streamline, isentropic, adiabatic, turbulence, friction

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INTRODUCTION

Various forms of Bernoulli's equation can be modeled because of the existence of

various types of fluid flow, and therefore Bernoulli's principle can be applied too.

Bernoulli's principle states that, an increase in the speed of a fluid flow occurs

simultaneously with a decrease in internal pressure. The principle is named after

Daniel Bernoulli published it in his book Hydrodynamic in 1738. Although, Bernoulli

deduced that pressure decreases when the flow speed increases, it was Leonhard

Euler who derived Bernoulli's equation in its usual form in 1752. All in all, there is a

correct way of using Bernoulli`s equation with confidence and which is briefly

described to continuation.

The law that explained the phenomenon from the energy conservation point of view

was found in his Hydrodynamic work. Later, Euler deduced an equation for an

inviscid flow (assuming that viscosity was insignificant) from which Bernoulli’s

equation arises naturally when considering a stationary case subjected to a

conservative gravitational field.

DISCUSSION

To arrive at Bernoulli`s equation, certain assumptions had to be made which limit us

the level of applicability. According to Euler equation Eq. 1, defined by Anderson, Jr.

(1989), it gives the variation of pressure with respect to speed variation, ignoring

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shear forces (inviscid fluid flow) and body forces (weight of the air fluid particle is

ignored). Only pressure forces were considered.

𝑑𝑝 = −𝜌𝑣𝑑𝑣 (1)

Integrating Eq. (1) by using a limit integration and considering an incompressible

fluid flow (change in density is very small because of low speed), will give us

Bernoulli`s equation applicable to points 1 and 2 which are on the same streamline.

However, if the flow is uniform throughout the field, then the constant in Eq. (2) is

the same for all streamlines as defined by Anderson, Jr. (1989).

∫𝑃1

𝑃2𝑑𝑝 = −𝜌∫

𝑉1

𝑉2𝑉𝑑𝑣

𝑃2 − 𝑃1 =−1

2𝜌(𝑉2

2 − 𝑉12)

𝑃1 +1

2𝜌𝑉1

2 = 𝑃2 +1

2𝜌𝑉2

2 = const along streamline

(2)

The assumptions that were made during the derivation of this equation led us to

some restrictions that must be implemented in order to use Bernoulli`s equation. But,

first of all, we must verify if the flow field in question is possible to exist. This is done

by verifying if Continuity Equation is fulfilled.

1. Continuity Equation in its Vector Form

The continuity equation states that, “the net outflow of mass through the surface

surrounding the volume must be equal to the decrease of mass within the volume”

(Bertin and Smith 1998, p. 24). This is, when a fluid is in motion, it must move in

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such a way that mass is conserved as it is stated in Eq. 3 defined by Bertin and

Smith (1998).

𝜕𝜌

𝜕𝑡+ 𝜵(𝜌𝑽

) = 0 (3)

Where ρ is the fluid density, t is the time, 𝑽→

is the flow velocity vector field

2. Steady Flow

To see further how mass conservation places restrictions on the velocity field,

consider a steady fluid flow. That is, for a relatively low speed flow, the pressure

variations are sufficiently small, and because of this, the density change is also small

that can be assumed to be constant and so, the density of the fluid flow does not

vary with time.

𝜕𝜌

𝜕𝑡= 0

3. Incompressible

Since density change is very small for low velocity airflows, it can be assumed to be

constant (𝜌 = 𝑐𝑜𝑛𝑠𝑡). One way to proof this, is by verifying if we are dealing with low

velocity airflows. As a rule of thumb, if its Mach Number is lower than 0,3 or has a

velocity less than 300 ft/s or 100m/s (or approximately 200 mph), then the velocity

airflow can be assumed to be small and treated as incompressible Anderson, (1989)

and Anderson (2003).

𝑀 =𝑉

𝑎=

𝑉

√𝛾𝑅𝑇

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Where:

V is the flow velocity

a is the Speed of Sound

T is the temperature of the flow field

γ is the ratio of the specific heats at constant pressure and volume respectively

and has a value of γ = 1.4 for dry air.

R is the air constant for ideal gas and has a value of 𝑅 = 1716𝑓𝑡.𝑙𝑏

𝑠𝑙𝑢𝑔𝑠.𝑅 𝑜𝑟 𝑅 =

287𝑁.𝑊

𝐾𝑔.𝐾

Then, so far, the continuity equation reduces to:

𝛻. 𝑽→

= 0 or:

𝜕𝑢

𝜕𝑥+

𝜕𝑣

𝜕𝑦+

𝜕𝑤

𝜕𝑧= 0 (4)

4. Remember that shear forces (friction) and body forces (gravity) were ignored

to get Eq. (1). But in the case where conservative body forces are considered like

gravity, Bernoulli`s equation would include an extra term ρgh as shown in Eq. (5).

In many applications of Bernoulli`s equation, the change in the term ρgh (change

in potential energy) along the streamline flow is so small in comparison to the other

terms that it can be ignored. For example, in the case of an aircraft in flight, the

change in height “h” along a streamline flow is so small that the term ρgh can be

disregarded.

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1

2𝜌𝑉2 + 𝑃 + 𝜌𝑔ℎ = 𝑐𝑜𝑛𝑠𝑡

(5)

1

2𝜌𝑉2 + 𝑃 = 𝑐𝑜𝑛𝑠𝑡 = Ptotal

where

Static pressure = P

Dynamic pressure = 1

2𝜌𝑉2

Total Pressure = Ptotal

5. Inviscid Flow: The product of viscosity times shear velocity gradient defines the

term shear stress, 𝜏. We must understand that, there are no real fluids for which

viscosity is zero. But, there are many real cases where this product is sufficiently

small that, the shear stress term, can be ignored when compared to other terms

in the governing equations as described by Bertin and Smith (1998).

6. Irrotational flow. If the 2D flow contains no singularities, then the Vorticity Vector

�⃗⃗⃗� in Eq. (6) for irrotational flow must be zero as defined by Bertin and Smith

(1998).

�⃗⃗⃗� = 𝜵 × 𝑽→

= (𝒊𝜕

𝜕𝑥+ 𝒋

𝜕

𝜕𝑦) × (𝒊𝑢 + 𝒋𝑣) = 𝟎

(6)

𝜵 × 𝑽→

= (𝜕𝑣

𝜕𝑥+

𝜕𝑢

𝜕𝑦)𝑘 = 0

or

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ω = 𝜕𝑣

𝜕𝑥−

𝜕𝑢

𝜕𝑦= 0

(7)

If ω = 0 (irrotational flow), then the constant in Eq. (2) is real in all the fluid flow. But

if ω≠0 (rotational flow), then, this constant is only real along a streamline. Here, we

present an example of the correct way of using Bernoulli`s equation.

Let`s consider a 2D velocity flow field at sea level (𝜌 = 1.225𝐾𝑔

𝑓𝑡3) and defined by:

𝑢 = 𝑥2 − 𝑥𝑦 (8)

𝑣 =𝑦2

2− 2𝑥𝑦 (9)

Where “u” and “v” are defined in m/s.

First, we need to identify if continuity equation is satisfied, or in other words, if the

velocity flow field is possible to exist.

Continuity equation in a 2D form is:

𝜕𝑢

𝜕𝑥+

𝜕𝑣

𝜕𝑦= 0 (10)

Then, 𝜕𝑢

𝜕𝑥= 2𝑥 − 𝑦

(11)

and 𝜕𝑣

𝜕𝑦= 𝑦 − 2𝑥

(12)

Substituting equations (11) and (12) into Eq. (10) yields:

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2𝑥 − 𝑦 + 𝑦 − 2𝑥 = 0

(13)

Eq. (13) shows that the given flow velocity field satisfy the continuity equation.

Second, we need to find out if the given flow velocity field is rotational or

irrotational.

Evaluating the partial derivatives of equations (8) and (9) yield equations (14) and

(15)

respectively.

ω =𝜕𝑣

𝜕𝑥−

𝜕𝑢

𝜕𝑦

𝜕𝑣

𝜕𝑥= −2𝑦 (14)

𝜕𝑢

𝜕𝑦= −𝑥 (15)

Substituting equations (14) and (15) into Eq. (7) yields:

ω = −2𝑦 − (−𝑥) = 𝑥 − 2𝑦 ≠ 0

It is clear that, the given velocity flow field is rotational (ω ≠ 0). So, that means that

we can still use Bernoulli`s eq. only if the two given points are on the same

streamline. So, we need to identify the streamline by using the 2D streamline Eq.

(16) defined by Bertin and Smith (1998):

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𝑑𝑥

𝑢=

𝑑𝑦

𝑣 (16)

Solving this equation for the given velocity components shown in equations (8)

and (9) one finds that

𝑑𝑥

𝑥2 − 𝑥𝑦=

𝑑𝑦

𝑦2

2 − 2𝑥𝑦

and

(𝑦2

2− 2𝑥𝑦 )𝑑𝑥 − (𝑥2 − 𝑥𝑦) 𝑑𝑦 = 0

Since this equation is a point function, then

𝑑𝜑 =𝜕𝐹

𝜕𝑥𝑑𝑥 −

𝜕𝐹

𝜕𝑦𝑑𝑦 = 0

where 𝜕𝐹

𝜕𝑥= 𝑣 =

𝑦2

2− 2𝑥𝑦

(17) and

𝜕𝐹

𝜕𝑦= −𝑢 = −(𝑥2 − 𝑥𝑦) (18)

Integrating Eq. (17) with respect to “x” yields

𝜑 = 𝐹 =𝑦2

2𝑥 − 𝑥2𝑦 + 𝑓(𝑦) (19)

Where 𝜑 = 𝐹 is the respective stream function Then, taking the derivate of Eq. (19) with respect to “y” yields

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𝜕𝐹

𝜕𝑦= 𝑦𝑥 − 𝑥2 + 𝑓′(𝑦) (20)

Replacing Eq. (20) into Eq. (18) yields

−𝑥2 + 𝑥𝑦 = 𝑦𝑥 − 𝑥2 + 𝑓′(𝑦)

𝑓′(𝑦) = 0

Therefore

𝑓(𝑦) = 𝑐𝑜𝑛𝑠𝑡

And the streamline is:

𝜑 =𝑦2

2𝑥 − 𝑥2𝑦 + 𝑐𝑜𝑛𝑠𝑡 = 0

(21)

Now, if we intend to use Bernoulli`s eq, for example to find the static pressure

difference between two points in the flow, we must be sure to have these two points

on the same streamline.

Consider these two points to be: (-1, 2) and (2, 2). The coordinates of these two

points are defined in meters.

For point one (-1, 2), Eq. (21) results in -4.

(2)2

2(−1) − (−1)2(2) = −4

For point two (2, 2), Eq. (21) also results in -4

(2)2

2(2) − (2)2(2) = −4

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Then, these two points are on the same streamline, so we can use Bernoulli`s

equation only between these two points even though the flow is rotational. Now,

using Eq. (8) and (9)

𝑢 = 𝑥2 − 𝑥𝑦

𝑣 =𝑦2

2− 2𝑥𝑦

For point one (-1, 2)

𝑢 = (−1)2 − (−1)(2) = 1- (-2) = 3 m/s

𝑣 =(𝑦)2

2− 2(𝑥)(𝑦) =

(−1)2

2− 2(−1)(2) =

1

2− 2(−2) =

1

2+ 4 =

9

2 = 4.5 m/s

𝑉1 = √(3)2 + (4.5)2 = 5.41 m/s

For point two (2, 2)

𝑢 = (2)2 − (2)(2) = 4 - 4 = 0 m/s

𝑣 =(2)2

2− 2(2)(2) = 2 – 8 = - 6 m/s (negative sign means opposite direction)

𝑉2 = √(0)2 + (−6)2 = 6 𝑚/𝑠

Using Bernoulli`s Eq. (2):

𝑃1 +1

2𝜌𝑉1

2 = 𝑃2 +1

2𝜌𝑉2

2

𝑃1 +1

2∗ 1.225 ∗ 5.412 = 𝑃2 +

1

2∗ 1.225 ∗ 62

𝑃1 − 𝑃2 = 4.123 𝑃𝑎

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The static pressure difference between 𝑃1 and 𝑃2 which are located on the same

streamline in the fluid flow is 4.123 Pa.

CONCLUSION

Throughout this paper, the correct way of using Bernoulli`s equation has been

shown. Initially, it has been presented some assumptions for which this equation is

valid to apply. These assumptions led to a set of restrictions that must be met in

order to apply correctly this equation. However, during the process of the application

of Bernoulli`s equation, the analyst has to be sure which restrictions apply for the

particular case. According to this, the results must be presented in a similar way as

it was done in this paper like the pressure difference between the two points on the

same streamline.

REFERENCES

Anderson, Jr, J.D. (1989). Introduction to Flight. 3rd ed. New York: McGraw-Hill.

Anderson, Jr. J.D. (2003). Modern Compressible Flow with Historical Perspective.

2nd ed. New York: McGraw-Hill.

Bertin, J.J. and Smith, M.L. (1998). Aerodynamics for Engineer. 3rd ed. New Jersey:

Prentice Hall.

Luis A. Arriola*

Facultad de Ingeniería y Arquitectura

Escuela Profesional de Ciencias Aeronáuticas

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Universidad de San Martín de Porres, Lima Perú

* Aerospace Engineer, B.S., M.S.

[email protected]